How Much Ice Melts When Shot with a Lead Bullet?

I am trying to find the amount of ice that melts when a lead bullet traveling at 240 m/s at 30 °C. This textbook claims I should use the formula
[tex] \dfrac{1}{2} m v^2 + m_{bullet} c_{bullet} \left|\Delta \right| T = L_f \Delta m[/tex]

What I don't understand is why I shouldn't first calculate how hot the bullet gets after embedding into the ice using
[tex] \dfrac{1}{2} m v^2 = mc \left( T_f-30^{\circ}C \right)[/tex]

Then calculate the amount of ice that melts using:
[tex] L \Delta m = -m_{bullet} c_{bullet} \left( T_f - T_i \right) [/tex]
Where Tf is 0°C and Ti is the Tf after the embedding in the ice.

The data is that the mass of the bullet is 3.00g, so (1/2)mv^2 = mc(Tf-Ti) gives a Tf of 481 deg Celsius, which would cause a change in mass (melting of the ice into water) of .56 g but this is incorrect. Can anyone explain why this method is wrong?